Skip to content
Related Articles
Count of triplets that satisfy the given equation
• Difficulty Level : Medium
• Last Updated : 04 Jun, 2021

Given an array arr[] of N non-negative integers. The task is to count the number of triplets (i, j, k) where 0 ≤ i < j ≤ k < N such that A[i] ^ A[i + 1] ^ … ^ A[j – 1] = A[j] ^ A[j + 1] ^ … ^ A[k] where ^ is the bitwise XOR.
Examples:

Input: arr[] = {2, 5, 6, 4, 2}
Output:
The valid triplets are (2, 3, 4) and (2, 4, 4).
Input: arr[] = {5, 2, 7}
Output:

Naive approach: Consider each and every triplet and check whether the xor of the required elements is equal or not.
Efficient approach: If arr[i] ^ arr[i + 1] ^ … ^ arr[j – 1] = arr[j] ^ arr[j + 1] ^ … ^ arr[k] then arr[i] ^ arr[i + 1] ^ … ^ arr[k] = 0 since X ^ X = 0. Now the problem gets reduced to finding the sub-arrays with XOR 0. But every such sub-array can have multiple such triplets i.e.

If arr[i] ^ arr[i + 1] ^ … ^ arr[k] = 0
then, (arr[i]) ^ (arr[i + 1] ^ … ^ arr[k]) = 0
and, arr[i] ^ (arr[i + 1]) ^ … ^ arr[k] = 0
arr[i] ^ arr[i + 1] ^ (arr[i + 2]) ^ … ^ arr[k] = 0
…
j can have any value from i + 1 to k without violating the required property.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the count``// of required triplets``int` `CountTriplets(``int``* arr, ``int` `n)``{``    ``int` `ans = 0;``    ``for` `(``int` `i = 0; i < n - 1; i++) {` `        ``// First element of the``        ``// current sub-array``        ``int` `first = arr[i];``        ``for` `(``int` `j = i + 1; j < n; j++) {` `            ``// XOR every element of``            ``// the current sub-array``            ``first ^= arr[j];` `            ``// If the XOR becomes 0 then``            ``// update the count of triplets``            ``if` `(first == 0)``                ``ans += (j - i);``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 2, 5, 6, 4, 2 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << CountTriplets(arr, n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG``{` `// Function to return the count``// of required triplets``static` `int` `CountTriplets(``int``[] arr, ``int` `n)``{``    ``int` `ans = ``0``;``    ``for` `(``int` `i = ``0``; i < n - ``1``; i++)``    ``{` `        ``// First element of the``        ``// current sub-array``        ``int` `first = arr[i];``        ``for` `(``int` `j = i + ``1``; j < n; j++)``        ``{` `            ``// XOR every element of``            ``// the current sub-array``            ``first ^= arr[j];` `            ``// If the XOR becomes 0 then``            ``// update the count of triplets``            ``if` `(first == ``0``)``                ``ans += (j - i);``        ``}``    ``}``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = {``2``, ``5``, ``6``, ``4``, ``2``};``    ``int` `n = arr.length;` `    ``System.out.println(CountTriplets(arr, n));``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation of the approach` `# Function to return the count``# of required triplets``def` `CountTriplets(arr, n):` `    ``ans ``=` `0``    ``for` `i ``in` `range``(n ``-` `1``):` `        ``# First element of the``        ``# current sub-array``        ``first ``=` `arr[i]``        ``for` `j ``in` `range``(i ``+` `1``, n):` `            ``# XOR every element of``            ``# the current sub-array``            ``first ^``=` `arr[j]` `            ``# If the XOR becomes 0 then``            ``# update the count of triplets``            ``if` `(first ``=``=` `0``):``                ``ans ``+``=` `(j ``-` `i)` `    ``return` `ans` `# Driver code``arr ``=` `[``2``, ``5``, ``6``, ``4``, ``2` `]``n ``=` `len``(arr)``print``(CountTriplets(arr, n))` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `    ``// Function to return the count``    ``// of required triplets``    ``static` `int` `CountTriplets(``int``[] arr, ``int` `n)``    ``{``        ``int` `ans = 0;``        ``for` `(``int` `i = 0; i < n - 1; i++)``        ``{``    ` `            ``// First element of the``            ``// current sub-array``            ``int` `first = arr[i];``            ``for` `(``int` `j = i + 1; j < n; j++)``            ``{``    ` `                ``// XOR every element of``                ``// the current sub-array``                ``first ^= arr[j];``    ` `                ``// If the XOR becomes 0 then``                ``// update the count of triplets``                ``if` `(first == 0)``                    ``ans += (j - i);``            ``}``        ``}``        ``return` `ans;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = {2, 5, 6, 4, 2};``        ``int` `n = arr.Length;``    ` `        ``Console.WriteLine(CountTriplets(arr, n));``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``
Output:
`2`

Time Complexity: O(n2)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer DSA Live Classes

My Personal Notes arrow_drop_up